Question

A stone of mass m = 1.05 kg is released from a height of h = 2.1 m into a pool of water. At a time of t = 1.83 s after hitting the surface of the water, the stone's velocity has decreased by 50%.
What is the magnitude of the average force the stone experiences, in newtons, during the time t?

Answer 1

Answer:

Explanation:

ignoring air resistance, the kinetic energy at water impact will equal the potential energy converted

½mv² = mgh

v = √(2gh)

v = √(2(9.81)2.1) = 6.4188... m/s

after impact, an impulse will result in a change of momentum.

There is a downward impulse due to gravity equal to the weight of the stone and an upward average force due to water resistance and buoyancy force.

FΔt = mΔv

(F - mg)Δt = m(vf - vi)

(F - mg) = m(vf - vi)/Δt

F = m(vf - vi)/Δt + mg

F = m((vf - vi)/Δt + g)

F = 1.05(((½(-6.4188) - -6.4188)/ 1.83) + 9.81)

F = 12.14198...

F = 12.1 N